Cover the $n$-disc irredundantly with $n+1$ open sets. Suppose that the $(n+1)$-fold intersection is empty. Then is some $n$-fold intersection empty?

Question

Cover the $n$-disc with $n+1$ open sets $D^n = U_0 \cup \dotsb \cup U_n$. Suppose that $U_0 \cap \dotsb \cap U_n = \emptyset$. Suppose moreover that the cover is irredundant in the sense that no proper subset of the $U_i$’s covers $D^n$.

Question: Then is some $n$-fold intersection empty? That is, after reordering do we have $U_1 \cap \cdots \cap U_n = \emptyset$?

I believe the answer is yes.

Notes:

• When $n = 1$, this follows from the fact that the disk is connected.

• In general, we can look at the Čech nerve of the cover. We have that the geometric realization is contractible: $\lvert\coprod_{i_0 < \dotsb < i_\bullet} U_{i_0} \cap \dotsb \cap U_{i_\bullet}\rvert \equiv \lvert D^n\rvert \equiv \ast$. Since the $(n+1)$-fold intersection is empty, there is a map from the geometric realization the geometric realization of the $(n+1)$-cube with top and bottom removed (i.e. what you’d get from the Čech nerve if all the intersections are a point except for the $(n+1)$-fold one which is empty), which has homotopy type $S^{n}$. I suspect this map is always essential if all of the $k$-fold intersections $U_{i_1} \cap \dotsb \cap U_{i_k}$ are nonempty for $k \leq n$, which would be enough to prove an affirmative answer to my question.

• When $n = 2$, I believe that the intersections $U_{i_0} \cap \dotsb \cap U_{i_k}$ must have the homotopy types of disjoint unions of wedges of circles (because they are open subsets of $D^2$). Then I think I can convince myself that the Čech nerve is homotopy equivalent to something you can build up one cell at a time, and that it must have a wedge summand of $S^1$ (more precisely, the map from the previous bullet splits). But this is a bit messy, and will get messier as the dimension increases.

Answer 1

Too long for a comment.

In the case $n=2$ and if the $U_i$ are connected, we can prove that $U_1\cap U_2\cap U_3$ is nonempty if the $U_i\cap U_j$ are nonempty: letting $p_1,p_2,p_3$ be in $U_2\cap U_3,U_1\cap U_3,U_1\cap U_2$ respectively, we can consider polygonal arcs $\gamma_1$ from $p_2$ to $p_3$ inside $U_1$ and similarly $\gamma_2,\gamma_3$.

We can assume the three arcs only intersect in $p_1,p_2,p_3$; if not, e.g. if $\gamma_1,\gamma_2$ intersect in some point $q_3$, then change $p_3$ to $q_3$ and repeat this process finitely many times, as the arcs are polygonal.

Once the three arcs only intersect in $p_1,p_2,p_3$, consider the triangle they form. Applying the following lemma finishes the proof.

Lemma: If a triangle $T$ with sides $l_1,l_2,l_3$ is covered by open sets $U_1,U_2,U_3\subseteq T$ with $l_i\subseteq U_i$, then $U_1\cap U_2\cap U_3$ is nonempty.

To prove it, first note that we can choose compact sets $K_1,K_2,K_3$ with $l_i\subseteq K_i\subseteq U_i$ for $i=1,2,3$ and $T=K_1\cup K_2\cup K_3$ (let $K_i=\{p\in U_i;d(p,T\setminus U_i)\geq\varepsilon\}$ for small enough $\varepsilon$).

Now for each $\varepsilon$, consider a triangulation of $T$ by triangles of diameter $<\varepsilon$, and color all vertices $p$ of the triangulation by a color $i\in\{1,2,3\}$ such that $p\in K_i$. We can assume that all the vertices in $l_i$ have color $i$, except the intersections $l_1\cap l_2$, $l_2\cap l_3$ and $l_3\cap l_1$, which can be colored $2,3,1$ respectively. Then Sperner's lemma guarantees a triangle in the triangulation with vertices of three distinct colours. Letting $\varepsilon\to0$ and taking an accumulation point of the resulting triangles, we obtain a point in $K_1\cap K_2\cap K_3\subseteq U_1\cap U_2\cap U_3$.

Answer 2

It is not true in general. Let $n=3$, and take a piece of a brick wall (thickness one brick) that is homeomorphic to a 3-disc. For example, just take a largish rectangular wall. There are bricks of four colours, in multiple rows. In each even row the bricks are alternately colours $a,b,a,b,\ldots$ and in each odd row the bricks are alternately colours $c,d,c,d,\ldots$. The four open sets are the bricks of a given colour together with some surrounding mortar. Because of the way that brick walls are built, each 3-fold intersection is non-empty but the 4-fold intersection is empty.

Note that each of the open sets has lots of components, as do all of the 2- and 3-fold intersections. For counterexamples you will need the open sets and their intersections to be far from being contractible, so that the disc that is their union is far from being the nerve of the covering.

This answer would work a lot better with a picture, but hopefully it is understandable as it is.

The pieces can be made to be connected if you want too. To connect the bricks of a given colour, just bore holes through to the nearest neighbours of the same colour being careful that the holes avoid the 3-fold intersections - you can even ensure that each hole passes through just one brick of a different colour - and fill each bore hole with the same colour as the bricks that it connects between. Do this for each of the four colours, where of course you also need to avoid the (thin) boreholes that you have already made for one of the other colours.

In this case the pieces and some of their intersections have non-trivial first homology and second homology, and this is what keeps the cover from being too similar to its nerve.